Question

A negative test charge of 2.0×10−8 C is placed in an electric field. The force on the test charge is 0.06 N. What is the electric field intensity at the location of the test charge?

3.0 x 10
4
N/C
3.0×10
−5
N/C
3.0 x 10
6
N/C
5.0 x 10
5
N/C

Answer 1

To find the electric field intensity (E) at the location of the test charge, we can use the formula that relates electric field intensity, force (F), and charge (q):

\[
E = \frac{F}{q}
\]

where:
- \( E \) is the electric field intensity (in N/C),
- \( F \) is the force on the charge (in N),
- \( q \) is the charge (in C).

Given:
- \( F = 0.06 \, \text{N} \)
- \( q = 2.0 \times 10^{-8} \, \text{C} \)

Now, substituting the values into the equation:

\[
E = \frac{0.06 \, \text{N}}{2.0 \times 10^{-8} \, \text{C}}
\]

Calculating this gives:

\[
E = \frac{0.06}{2.0 \times 10^{-8}} = 3.0 \times 10^{6} \, \text{N/C}
\]

So, the electric field intensity at the location of the test charge is:

\[
\boxed{3.0 \times 10^{6} \, \text{N/C}}
\]